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We know that gauge theory is renormalizable, due to the Ward-Takahashi identity (for non-Abelian theory, it is Slavnov-Taylor identity), which reflects the conserved current of gauge symmetry.

But local (gauge) symmetry is not a real 'symmetry', since it cannot lead to a physical conserved current. When the gauge group is non-Abelian, local gauge invariance can lead to either a gauge-invariant but non-conserved current, or a gauge-dependent but conserved current (for $U(1)$ group these two currents coincide). But global symmetry, leads to a physical (global) invariant conserved current (for non-Abelian group, gauge field transform under global transformation, too), and this can lead to corresponding Ward-Takahashi identity.

Now here is my question, if a gauge theory is global but not local invariant, does it renormalizable? Specifically, if in SM Lagrangian we change Higgs covariant differential $D_\mu$ to ordinary differential $\partial_\mu$, does the theory renormalizable? If the change is done, then the Yukawa interaction term destroys the local $SU(2)\times U(1)$ symmetry, but conserves the global one.

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Gauge invariance is always a local symmetry. In this sense a gauge symmetry and a local symmetry can be taken as being synonymous. So there is no such thing as a global gauge symmetry. The local (gauge) symmetry is a real symmetry, because gauge transformations leave the Lagrangian invariant. One can also derive a conserved current (Noether current) for it, but one needs to use a bit of a trick.* One only allows the gauge field to be transformed and not the fermion fields. The resulting conserved current is then expressed in terms of the fermion fields. So, one can see that in an interacting gauge theory the gauge field couples to this conserved current.

So, if you were to change the local gauge symmetry into a global symmetry, you would not have the interactions anymore, because the gauge derivatives, which contain the interaction would disappear. The resulting theory would then decouple into two free-field theories, one for the gauge field and one for the fermion fields. Each of these would be trivially renormalizable, because there are no interactions.

*See, for example: M. E. Peskin and D. V. Schroeder, An Introduction to Quantum Field Theory, Addison Wesley (1995), Chapter 9.

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  • $\begingroup$ I'm not sure this really answers the question. I think the OP understands the local vs global distinction (although the OP's terminology of local gauge symmetry is poorly chosen) $\endgroup$ – innisfree Oct 3 '16 at 5:29
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    $\begingroup$ Could be that I misunderstood the question, but it seems to me the OP has some misconceptions about global symmetries, which I'm trying to clear up. $\endgroup$ – flippiefanus Oct 3 '16 at 5:33
  • $\begingroup$ Thanks for your answer. Yes, 'gauge' means 'local', and I made a mistake in the previous question. Now I've modified it. $\endgroup$ – Doggy Oct 3 '16 at 8:19
  • $\begingroup$ And the question I have is, if there is a covariant differential $D_\mu$ for fermion, but a ordinary differential $\partial_\mu$ for Higgs, will this Lagrangian renormalizable? The model is invariant under global $SU(2)\times U(1)$ symmetry, but non-invariant under the local one (Yukawa term breaks the symmetry). $\endgroup$ – Doggy Oct 3 '16 at 8:22
  • $\begingroup$ Oh dear. I think the issue is the mass of the gauge boson. Usually this mass is protected by the gauge symmetry. I'm not sure if this still applies when the symmetry is broken down to a global symmetry. If the mass is not protected anymore, it can become nonzero and then I believe the theory would not be renormalizable anymore, but I'm not perfectly sure. I'll think about it. $\endgroup$ – flippiefanus Oct 3 '16 at 10:07

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